//package Jama;
package no.uib.jexpress_modularized.pca.computation.Jama;

import no.uib.jexpress_modularized.pca.computation.Jama.util.*;



   /** Singular Value Decomposition.

   <P>

   For an m-by-n matrix A with m >= n, the singular value decomposition is

   an m-by-n orthogonal matrix U, an n-by-n diagonal matrix S, and

   an n-by-n orthogonal matrix V so that A = U*S*V'.

   <P>

   The singular values, sigma[k] = S[k][k], are ordered so that

   sigma[0] >= sigma[1] >= ... >= sigma[n-1].

   <P>

   The singular value decompostion always exists, so the constructor will

   never fail.  The matrix condition number and the effective numerical

   rank can be computed from this decomposition.

   */



public class SingularValueDecomposition implements java.io.Serializable {



/* ------------------------

   Class variables

 * ------------------------ */



   /** Arrays for internal storage of U and V.

   @serial internal storage of U.

   @serial internal storage of V.

   */

   private double[][] U, V;



   /** Array for internal storage of singular values.

   @serial internal storage of singular values.

   */

   private double[] s;



   /** Row and column dimensions.

   @serial row dimension.

   @serial column dimension.

   */

   private int m, n;



/* ------------------------

   Constructor

 * ------------------------ */



   /** Construct the singular value decomposition

   @param A    Rectangular matrix

   @return     Structure to access U, S and V.

   */



   public SingularValueDecomposition (Matrix Arg) {



      // Derived from LINPACK code.

      // Initialize.

      double[][] A = Arg.getArrayCopy();

      m = Arg.getRowDimension();

      n = Arg.getColumnDimension();

      int nu = Math.min(m,n);

      s = new double [Math.min(m+1,n)];

      U = new double [m][nu];

      V = new double [n][n];

      double[] e = new double [n];

      double[] work = new double [m];

      boolean wantu = true;

      boolean wantv = true;



      // Reduce A to bidiagonal form, storing the diagonal elements

      // in s and the super-diagonal elements in e.



      int nct = Math.min(m-1,n);

      int nrt = Math.max(0,Math.min(n-2,m));

      for (int k = 0; k < Math.max(nct,nrt); k++) {

         if (k < nct) {



            // Compute the transformation for the k-th column and

            // place the k-th diagonal in s[k].

            // Compute 2-norm of k-th column without under/overflow.

            s[k] = 0;

            for (int i = k; i < m; i++) {

               s[k] = Maths.hypot(s[k],A[i][k]);

            }

            if (s[k] != 0.0) {

               if (A[k][k] < 0.0) {

                  s[k] = -s[k];

               }

               for (int i = k; i < m; i++) {

                  A[i][k] /= s[k];

               }

               A[k][k] += 1.0;

            }

            s[k] = -s[k];

         }

         for (int j = k+1; j < n; j++) {

            if ((k < nct) & (s[k] != 0.0))  {



            // Apply the transformation.



               double t = 0;

               for (int i = k; i < m; i++) {

                  t += A[i][k]*A[i][j];

               }

               t = -t/A[k][k];

               for (int i = k; i < m; i++) {

                  A[i][j] += t*A[i][k];

               }

            }



            // Place the k-th row of A into e for the

            // subsequent calculation of the row transformation.



            e[j] = A[k][j];

         }

         if (wantu & (k < nct)) {



            // Place the transformation in U for subsequent back

            // multiplication.



            for (int i = k; i < m; i++) {

               U[i][k] = A[i][k];

            }

         }

         if (k < nrt) {



            // Compute the k-th row transformation and place the

            // k-th super-diagonal in e[k].

            // Compute 2-norm without under/overflow.

            e[k] = 0;

            for (int i = k+1; i < n; i++) {

               e[k] = Maths.hypot(e[k],e[i]);

            }

            if (e[k] != 0.0) {

               if (e[k+1] < 0.0) {

                  e[k] = -e[k];

               }

               for (int i = k+1; i < n; i++) {

                  e[i] /= e[k];

               }

               e[k+1] += 1.0;

            }

            e[k] = -e[k];

            if ((k+1 < m) & (e[k] != 0.0)) {



            // Apply the transformation.



               for (int i = k+1; i < m; i++) {

                  work[i] = 0.0;

               }

               for (int j = k+1; j < n; j++) {

                  for (int i = k+1; i < m; i++) {

                     work[i] += e[j]*A[i][j];

                  }

               }

               for (int j = k+1; j < n; j++) {

                  double t = -e[j]/e[k+1];

                  for (int i = k+1; i < m; i++) {

                     A[i][j] += t*work[i];

                  }

               }

            }

            if (wantv) {



            // Place the transformation in V for subsequent

            // back multiplication.



               for (int i = k+1; i < n; i++) {

                  V[i][k] = e[i];

               }

            }

         }

      }



      // Set up the final bidiagonal matrix or order p.



      int p = Math.min(n,m+1);

      if (nct < n) {

         s[nct] = A[nct][nct];

      }

      if (m < p) {

         s[p-1] = 0.0;

      }

      if (nrt+1 < p) {

         e[nrt] = A[nrt][p-1];

      }

      e[p-1] = 0.0;



      // If required, generate U.



      if (wantu) {

         for (int j = nct; j < nu; j++) {

            for (int i = 0; i < m; i++) {

               U[i][j] = 0.0;

            }

            U[j][j] = 1.0;

         }

         for (int k = nct-1; k >= 0; k--) {

            if (s[k] != 0.0) {

               for (int j = k+1; j < nu; j++) {

                  double t = 0;

                  for (int i = k; i < m; i++) {

                     t += U[i][k]*U[i][j];

                  }

                  t = -t/U[k][k];

                  for (int i = k; i < m; i++) {

                     U[i][j] += t*U[i][k];

                  }

               }

               for (int i = k; i < m; i++ ) {

                  U[i][k] = -U[i][k];

               }

               U[k][k] = 1.0 + U[k][k];

               for (int i = 0; i < k-1; i++) {

                  U[i][k] = 0.0;

               }

            } else {

               for (int i = 0; i < m; i++) {

                  U[i][k] = 0.0;

               }

               U[k][k] = 1.0;

            }

         }

      }



      // If required, generate V.



      if (wantv) {

         for (int k = n-1; k >= 0; k--) {

            if ((k < nrt) & (e[k] != 0.0)) {

               for (int j = k+1; j < nu; j++) {

                  double t = 0;

                  for (int i = k+1; i < n; i++) {

                     t += V[i][k]*V[i][j];

                  }

                  t = -t/V[k+1][k];

                  for (int i = k+1; i < n; i++) {

                     V[i][j] += t*V[i][k];

                  }

               }

            }

            for (int i = 0; i < n; i++) {

               V[i][k] = 0.0;

            }

            V[k][k] = 1.0;

         }

      }



      // Main iteration loop for the singular values.



      int pp = p-1;

      int iter = 0;

      double eps = Math.pow(2.0,-52.0);

      while (p > 0) {

         int k,kase;



         // Here is where a test for too many iterations would go.



         // This section of the program inspects for

         // negligible elements in the s and e arrays.  On

         // completion the variables kase and k are set as follows.



         // kase = 1     if s(p) and e[k-1] are negligible and k<p

         // kase = 2     if s(k) is negligible and k<p

         // kase = 3     if e[k-1] is negligible, k<p, and

         //              s(k), ..., s(p) are not negligible (qr step).

         // kase = 4     if e(p-1) is negligible (convergence).



         for (k = p-2; k >= -1; k--) {

            if (k == -1) {

               break;

            }

            if (Math.abs(e[k]) <= eps*(Math.abs(s[k]) + Math.abs(s[k+1]))) {

               e[k] = 0.0;

               break;

            }

         }

         if (k == p-2) {

            kase = 4;

         } else {

            int ks;

            for (ks = p-1; ks >= k; ks--) {

               if (ks == k) {

                  break;

               }

               double t = (ks != p ? Math.abs(e[ks]) : 0.) + 

                          (ks != k+1 ? Math.abs(e[ks-1]) : 0.);

               if (Math.abs(s[ks]) <= eps*t)  {

                  s[ks] = 0.0;

                  break;

               }

            }

            if (ks == k) {

               kase = 3;

            } else if (ks == p-1) {

               kase = 1;

            } else {

               kase = 2;

               k = ks;

            }

         }

         k++;



         // Perform the task indicated by kase.



         switch (kase) {



            // Deflate negligible s(p).



            case 1: {

               double f = e[p-2];

               e[p-2] = 0.0;

               for (int j = p-2; j >= k; j--) {

                  double t = Maths.hypot(s[j],f);

                  double cs = s[j]/t;

                  double sn = f/t;

                  s[j] = t;

                  if (j != k) {

                     f = -sn*e[j-1];

                     e[j-1] = cs*e[j-1];

                  }

                  if (wantv) {

                     for (int i = 0; i < n; i++) {

                        t = cs*V[i][j] + sn*V[i][p-1];

                        V[i][p-1] = -sn*V[i][j] + cs*V[i][p-1];

                        V[i][j] = t;

                     }

                  }

               }

            }

            break;



            // Split at negligible s(k).



            case 2: {

               double f = e[k-1];

               e[k-1] = 0.0;

               for (int j = k; j < p; j++) {

                  double t = Maths.hypot(s[j],f);

                  double cs = s[j]/t;

                  double sn = f/t;

                  s[j] = t;

                  f = -sn*e[j];

                  e[j] = cs*e[j];

                  if (wantu) {

                     for (int i = 0; i < m; i++) {

                        t = cs*U[i][j] + sn*U[i][k-1];

                        U[i][k-1] = -sn*U[i][j] + cs*U[i][k-1];

                        U[i][j] = t;

                     }

                  }

               }

            }

            break;



            // Perform one qr step.



            case 3: {



               // Calculate the shift.

   

               double scale = Math.max(Math.max(Math.max(Math.max(

                       Math.abs(s[p-1]),Math.abs(s[p-2])),Math.abs(e[p-2])), 

                       Math.abs(s[k])),Math.abs(e[k]));

               double sp = s[p-1]/scale;

               double spm1 = s[p-2]/scale;

               double epm1 = e[p-2]/scale;

               double sk = s[k]/scale;

               double ek = e[k]/scale;

               double b = ((spm1 + sp)*(spm1 - sp) + epm1*epm1)/2.0;

               double c = (sp*epm1)*(sp*epm1);

               double shift = 0.0;

               if ((b != 0.0) | (c != 0.0)) {

                  shift = Math.sqrt(b*b + c);

                  if (b < 0.0) {

                     shift = -shift;

                  }

                  shift = c/(b + shift);

               }

               double f = (sk + sp)*(sk - sp) + shift;

               double g = sk*ek;

   

               // Chase zeros.

   

               for (int j = k; j < p-1; j++) {

                  double t = Maths.hypot(f,g);

                  double cs = f/t;

                  double sn = g/t;

                  if (j != k) {

                     e[j-1] = t;

                  }

                  f = cs*s[j] + sn*e[j];

                  e[j] = cs*e[j] - sn*s[j];

                  g = sn*s[j+1];

                  s[j+1] = cs*s[j+1];

                  if (wantv) {

                     for (int i = 0; i < n; i++) {

                        t = cs*V[i][j] + sn*V[i][j+1];

                        V[i][j+1] = -sn*V[i][j] + cs*V[i][j+1];

                        V[i][j] = t;

                     }

                  }

                  t = Maths.hypot(f,g);

                  cs = f/t;

                  sn = g/t;

                  s[j] = t;

                  f = cs*e[j] + sn*s[j+1];

                  s[j+1] = -sn*e[j] + cs*s[j+1];

                  g = sn*e[j+1];

                  e[j+1] = cs*e[j+1];

                  if (wantu && (j < m-1)) {

                     for (int i = 0; i < m; i++) {

                        t = cs*U[i][j] + sn*U[i][j+1];

                        U[i][j+1] = -sn*U[i][j] + cs*U[i][j+1];

                        U[i][j] = t;

                     }

                  }

               }

               e[p-2] = f;

               iter = iter + 1;

            }

            break;



            // Convergence.



            case 4: {



               // Make the singular values positive.

   

               if (s[k] <= 0.0) {

                  s[k] = (s[k] < 0.0 ? -s[k] : 0.0);

                  if (wantv) {

                     for (int i = 0; i <= pp; i++) {

                        V[i][k] = -V[i][k];

                     }

                  }

               }

   

               // Order the singular values.

   

               while (k < pp) {

                  if (s[k] >= s[k+1]) {

                     break;

                  }

                  double t = s[k];

                  s[k] = s[k+1];

                  s[k+1] = t;

                  if (wantv && (k < n-1)) {

                     for (int i = 0; i < n; i++) {

                        t = V[i][k+1]; V[i][k+1] = V[i][k]; V[i][k] = t;

                     }

                  }

                  if (wantu && (k < m-1)) {

                     for (int i = 0; i < m; i++) {

                        t = U[i][k+1]; U[i][k+1] = U[i][k]; U[i][k] = t;

                     }

                  }

                  k++;

               }

               iter = 0;

               p--;

            }

            break;

         }

      }

   }



/* ------------------------

   Public Methods

 * ------------------------ */



   /** Return the left singular vectors

   @return     U

   */



   public Matrix getU () {

      return new Matrix(U,m,Math.min(m+1,n));

   }



   /** Return the right singular vectors

   @return     V

   */



   public Matrix getV () {

      return new Matrix(V,n,n);

   }



   /** Return the one-dimensional array of singular values

   @return     diagonal of S.

   */



   public double[] getSingularValues () {

      return s;

   }



   /** Return the diagonal matrix of singular values

   @return     S

   */



   public Matrix getS () {

      Matrix X = new Matrix(n,n);

      double[][] S = X.getArray();

      for (int i = 0; i < n; i++) {

         for (int j = 0; j < n; j++) {

            S[i][j] = 0.0;

         }

         S[i][i] = this.s[i];

      }

      return X;

   }



   /** Two norm

   @return     max(S)

   */



   public double norm2 () {

      return s[0];

   }



   /** Two norm condition number

   @return     max(S)/min(S)

   */



   public double cond () {

      return s[0]/s[Math.min(m,n)-1];

   }



   /** Effective numerical matrix rank

   @return     Number of nonnegligible singular values.

   */



   public int rank () {

      double eps = Math.pow(2.0,-52.0);

      double tol = Math.max(m,n)*s[0]*eps;

      int r = 0;

      for (int i = 0; i < s.length; i++) {

         if (s[i] > tol) {

            r++;

         }

      }

      return r;

   }

}

